Calculate the pH in the titration of 50.0 mL of a 3.0 M H3PO3 (aq) with 100.0 mL

Question

Calculate the pH in the titration of 50.0 mL of a 3.0 M H3PO3 (aq) with 100.0 mL 3.0 M KOH (aq).

(*I am lost as H3PO3 is weak and diprotic how does this play into the amount of KOH needed and the combined total volume? Does the Molarity of the H3PO3 reduce to 1M and how much H3O+ does it contribute?)

(*can you please explain the answer and show your work, thanks)

(b) after addition of 25.0 mL of KOH 1.3 Number (c) after addition of 50.0 mL of KOH 4 Number (d) after addition of 75.0 mL of KOH 6.7 Number (aq), is a diprotic oxyacid that is an important compound in the industry and agriculture. Calculate the pH in the titration of 50.0 mL of a 3.0 M H3PO3 (aq) with 100.0 mL 3.0 M KOH (aq).

Explanation / Answer

first things first,

lets calculate the moles involved in the acid

M1*V1 = 50 ml*3M = 150 mmol of H3PO3 = 3*150 = 450 mmol of H+ possible

M2*V2 = 100 ml*3M = 300 mmol of KOH = 300 mmol of OH-

We will have an ACIDIC Solution

H+ and OH- will react all together to form water

then the solution here will be 50 ml + 100 ml = 150 ml in total (we will spend all the volume of KOH of course)

the left over of H3PO4 = 450 mmol H+ - 300 mmol = 150 mmol of H3PO4

or

50 mmol of H3PO4 are left...

now recalculate the concentrtaion of the acid

M3*V3 = mmol 3

M3 = 50 mmol / 150 ml = 0.333 M

[H3PO4] = 0.333 M

Now its time to start using

Ka1= 7.5 * 10^-3

Ka2= 6.2 * 10^-8

Ka3= 4.8 * 10^-13

NOTE = Ka1 >>> Ka2 and Ka3 ... you may assume that 98% of the H+ dissolved will be the first H+ and you may ignore the second and third hydrogens

Ka1 = [H+][H2PO4--]/[H3PO4]

since [H+]=[H2PO4--] lets assign it to x = [H+]=[H2PO4--]

[H3PO4] = 0.333 -x ... this is the actual molarity of the acid, minus the acid disolved which will be little... you may even cancel this "-x" value if you want... I will not so you may check the final value

7.5 * 10^-3 = x*x / (0.333 -x )

x^2 = (0.333-x)*(7.5 * 10^-3)

x^2 = 0.0025-(7.5 * 10^-3)x

x^2 + 7.5 * 10^-3 x -0.0025 = 0

Solve for x with quadratic formula

x = 0.0463 and -0.053

there are no negative concentrations so use 0.0463

Now that [H+] = 0.0463

you may calculate pH = -log(0.0463) = 1.33

pH = 1.33

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